Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.9 - Related Rates - 3.9 Exercises - Page 250: 29

Answer

The height is increasing at a rate of $~0.38~ft/min$

Work Step by Step

We can write an expression for the radius $r$: $2r = h$ $r = \frac{h}{2}$ We can differentiate both sides of the equation for volume with respect to $t$: $V = \frac{1}{3}\pi~r^2~h$ $V = \frac{1}{12}\pi~h^3$ $\frac{dV}{dt} = \frac{1}{4}\pi~h^2~\frac{dh}{dt}$ $\frac{dh}{dt} = \frac{4}{\pi~h^2}~\frac{dV}{dt}$ $\frac{dh}{dt} = [\frac{4}{\pi~(10~ft)^2}~]~(30~ft^3/min)$ $\frac{dh}{dt} = 0.38~ft/min$ The height is increasing at a rate of $~0.38~ft/min$
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